Two point charges 5×10^-8 culomb and …

Two charges q_A = 5 $\times$ 10^-8C and q_B = -3 $\times$ 10^-8C
Distance between two charges, r = 16 cm = 0.16 cm
Consider a point O on the line joining two charges where the electric potential is zero due to two charges.

From the figure we can see that, x = distance of point O from charge q_A
Electric potential at point O due to q_A,
$\mathrm{V}_{\mathrm{A}}=\frac{\mathrm{q}_{\mathrm{A}}}{4 \pi \varepsilon_{0}(\mathrm{AO})}$
$=9 \times 10^{9} \times \frac{5 \times 10^{-8}}{\mathrm{x}}$
$=\frac{450}{\mathrm{x}}$
Electric potential at point O due to q_B
$\mathrm{V}_{\mathrm{B}}=\frac{\mathrm{q}_{\mathrm{B}}}{4 \pi \varepsilon_{0}(\mathrm{BO})}$
$=9 \times 10^{9} \times \frac{-3 \times 10^{-8}}{0.16-\mathrm{x}}$
$=\frac{-270}{0.16-X}$
Since the total electric potential at O is zero,
$\Rightarrow$ V_A + V_B = 0
$\Rightarrow \frac{450}{x}+\left(-\frac{270}{0.16-x}\right)=0$
$\Rightarrow \frac{450}{x}=\frac{270}{0.16-x}$
$\Rightarrow \frac{5}{x}=\frac{3}{0.16-x}$
On cross multiplying we get,
5 $\times$ (0.16 - x) = 3x
$\Rightarrow$ = 0.8 - 5x = 3x
$\Rightarrow$ 8x = 0.8
$\Rightarrow$ x = 0.1m = 10cm (from charge q_A)
$\therefore$ at a distance of 10cm from the positive charge, the potential is zero between the two charges.